Let $\alpha ,\beta$ be the roots of the equation $x^2-\sqrt{2}x+\sqrt{6}=0$ and $\frac{1}{{\alpha }^2}+1,\frac{1}{{\beta }^2}+1$ be  the roots of the equation $x^2+ax+b=0$. Then the roots of the equation $x^2-\left(a+b-2\right)x+\left(a+b+2\right)=0$ are

 

$$\begin{gathered} a=\frac{-1}{{\alpha }^2}-\frac{1}{{\beta }^2}-2, \\ b=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+1+\frac{1}{{\alpha }^2{\beta }^2} \\ a+b=\frac{1}{{\left(\alpha \beta \right)}^2}-1=\frac{1}{6}-1=-\frac{5}{6} \\ {x}^2-\left(-\frac{5}{6}-2\right)x+\left(2-\frac{5}{6}\right)=0 \\ \Rightarrow 6x^2+17x+7=0 \\ \Rightarrow 6x^2+3x+14x+7=0 \\ \Rightarrow 3x(2x+1)+7(2x+1)=0 \\ \Rightarrow x=-\frac{7}{3}, x=-\frac{1}{2} are the roots \\ \therefore Both roots are real and negative. \end{gathered}$$